Solving RMQ (range of most value) problem with the ST algorithm (ST table)

// concept 3264

//O(n log(n) )

// can only be used for static data and are only seeking the maximum or minimum interval

. 1 #include <cstdio>
 2 #include <the cmath>
 . 3 #include <the iostream>
 . 4 #include <algorithm>
 . 5  the using  namespace STD;
 . 6  
. 7  const  int MAXN = 5E4 + . 5 ;
 . 8  
. 9  int n, m; // number n m queries 
10  int a [MAXN];
 . 11  int MAXN [MAXN] [ 32 ]; // F [i] [J] represents the number 2 ^ j starting from the maximum value of i-th digit interval [i, i + (<<. 1 J) -1] 
12 is  int Minn [MAXN] [ 32 ];
 13 is  
14  void ST () {
15     for(int i = 1; i <= n; ++i) maxn[i][0] = minn[i][0] = a[i];
16     int m = (int)log2(n*1.0)/log2(2*1.0);
17     for(int j = 1; j <= m; ++j) {
18         for(int i = 1; i+(1<<j)-1 <= n; ++i) {
19             maxn[i][j] = max(maxn[i][j-1], maxn[i+(1<< (J- 1 ))] [J- 1 ]);
20              by [i] [j] = min (a [i] [J- 1 ], a [i + ( 1 << (J- 1 ))] [J- 1 ]);
21          }
 22      }
 23  }
 24  
25  you query ( you have to, you r) {
 26      you k = log 2 (r-l + 1 ) / log 2 ( 2 );
27      return max (maxn [i] [k], maxn [R- ( 1 << k) + 1 ] [a]) - who (by [] [k], from [R- ( 1 << k) + 1][k]);
28 }
29 
30 int main() {
31     scanf("%d%d", &n, &m);
32     for(int i = 1; i <= n; ++i) {
33         scanf("%d", &a[i]);
34     }
35     ST();
36     int l, r;
37     for(int i = 0; i != m; ++i) {
38         scanf("%d%d", &l, &r);
39         printf("%d\n", query(l, r));
40     }
41     return 0;
42 }

Detailed // See https://blog.csdn.net/a_bright_ch/article/details/81062039